Here was the original problem statement.
Enumerate all non-isomorphic groups of order $175$.
See that $|G| = 175 = 5^2\cdot7$. Therefore by Sylow's first $H \leq G$ & $|H| = 25$ in addition to $K \leq G$ & $|K| = 7$. By Sylow's third $n_5 \equiv 1\pmod{5}$ and $n | 7$, so $n_5 = 1$ and $H$ is normal. By Sylow's third $n_7 \equiv 1\pmod{7}$ and $n| 25$, so $n_7 = 1$ and $H$ is normal. So both $H$ and $K$ are normal subgroups. We also know that $H \cap K = \{e\}$ because $\gcd(5, 7) = 1$.
Because $H \lhd G$ or $K \lhd G \Rightarrow HK \leq G$, therefore $HK \leq G$. $|HK| = 25 \cdot 7$ so $HK \cong G$. Because both groups are normal, $G \cong HK \cong H \times K$. Because $|H| = p^2 = 5^2$, therefore $G \cong \mathbb{Z}_5 \times \mathbb{Z}_5 \times \mathbb{Z}_7$ or $G \cong \mathbb{Z}_{25} \times \mathbb{Z}_7$.
But my proof is incomplete. I said that $G$ must be isomorphic to either $\mathbb{Z}_5 \times \mathbb{Z}_5 \times \mathbb{Z}_7$ or $\mathbb{Z}_{25} \times \mathbb{Z}_7$… but it could be possible that all groups of order 175 are isomorphic to only $\mathbb{Z}_{25} \times \mathbb{Z}_7$ and never $\mathbb{Z}_5 \times \mathbb{Z}_5 \times \mathbb{Z}_7$ . So I haven't shown that the group $\mathbb{Z}_5 \times \mathbb{Z}_5 \times \mathbb{Z}_7$ isn't an extraneous solution! How do I show that these groups are non-distinct?
Best Answer
You took a group of order $175$ (without making any other assumptions about the group) and you proved that it must be isomorphic to one of these two groups. So I don't really understand the question. $\mathbb{Z_{175}}$ is isomorphic to $\mathbb{Z_{25}}\times\mathbb{Z_7}$, it follows from the Chinese remainder theorem.